Applications-Of-Derivatives Question 446
Question: The line $ x+y=2 $ is tangent to the curve $ x^{2}=3-2y $ at its point
[MP PET 1998]
Options:
A) (1, 1)
B) (?1, 1)
C) ( $ \sqrt{3} $ , 0)
D) (3, ?3)
Show Answer
Answer:
Correct Answer: A
Solution:
Given curve $ x^{2}=3-2y $ …(i) Differentiate w.r.t. x, $ 2x=0-2\frac{dy}{dx} $
$ \Rightarrow $ $ \frac{dy}{dx}=-x $ Slope of the tangent of the curve $ =-x $ From the given line, slope $ =-1 $ ,
$ \therefore x=1 $ and from equation (i), $ y=1 $ .
$ \therefore $ Co-ordinate of the point is (1, 1).
BETA
